TITLE: Why are singular values always non-negative?
QUESTION [20 upvotes]: I have read that the singular values of any matrix $A$ are non-negative (e.g. wikipedia).
Is there a reason why?
The first possible step to get the SVD of a matrix $A$ is to compute $A^{T}A$. Then the singular values are the square root of the eigenvalues of $A^{T}A$. The matrix $A^{T}A$ is a symmetric matrix for sure. The eigenvalues of symmetric matrices are always real. But why are the eigenvalues (or the singular values) in this case always non-negative as well?

REPLY [8 votes]: Assume that $A$ is real for simplicity. The set of (orthogonal,diagonal,orthognal) matrices $(U, \Sigma, V)$ such that $A = U \Sigma V^T$ is not unique. Indeed, if $A = U \Sigma V^T$ then also 
$$ A = (-U)(-\Sigma)V^T = U (-\Sigma)(-V^T) = (UD_1)(D_1 \Sigma D_2)(V D_2)^T$$
for any diagonal matrices $D_1$ and $D_2$ with only $1$ or $-1$ on the diagonal.
Therefore, the positivity of the singular values is purely conventional.